Finding extremum of quadratic function using factorisation as symmetry operation.

Question

In this book in Section 3.2.3, the author shows how to find extremum of a quadratic function using an invariant quality and symmetry operation, which preserves it.

E.g. to find the extremum of $f(x)=6x-x^2$ we can notice that $f(x)=6x-x^2 = x(6-x) = (6-x)x$. This operation maps every value $x$ to $6-x$ through the axis of symmetry and vice-versa (e.g. 0 is mapped to 6 and 6 mapped to 0). This operation preserves symmetry for $x=3$ - the axis of symmetry and hence the extremum of the function.

What about the function $f(x)=6x+x^2 = x(6+x) = (6+x)x$? We know that the minimum of this function is at $x=-3$ but $6+(-3)=3$. Also this operation maps e.g $4$ to $10$, but $10$ to $16$! Hence, this approach fails with this example (symmetry is not preserved).

Why is that so? Why is this approach not working for all the quadratic functions?

Answer 1

It will work, but you need to find the correct symmetry transformation. Let's talk more about the first example, $f(x) = x(6-x)$. The symmetry transformation here is $x\mapsto 6-x$ because if we replace every $x$ in $x(6-x)$ with $6-x$ we get $$(6-x)(6-(6-x)) = (6-x)(6-6+x) = (6-x)x = x(6-x) = f(x).$$ In other words, $f(6-x) = f(x)$. We do the transformation, in this case replace $x$ with $6-x$, and arrive at the original function.

Now for your second example, $f(x) = x(6+x)$. To use the same argument we need to find the right transformation. In this case, the transformation is $x\mapsto -x-6$. We can verify this: $$f(-x-6) = (-x-6)(6-x-6) = (-x-6)(-x) = (x+6)x = x(6+x) = f(x).$$ Furthermore, the transformation $x\mapsto -x-6$ maps $4$ to $-10$ and $-10$ to $4$, and we can verify symmetry in general as well. In this case the only value that doesn't change is $x = -3$, because $-(-3) - 6 = -3$. So, the axis of symmetry, as well as the minimum of the function, is at $x = -3$.

Answer 2

$4$ doesn't get mapped to $10$. $4$ gets mapped to $-10$. And $-10$ gets mapped to $4$. It works.

The axis of symmetry is $x=-3$. So if you have a point $s = -3 + M$ it will be mapped to $-3 -M$.

So $4 = -3+M$ so $M = 7$. So $4 \mapsto -3 - 7 = -10$. And $-10 = -3 + M'$ so $M' = -7$ so $-10 \mapsto -3 -(-7) = 4$. Everything is good.

More generally if $x = -3 + M$ then $M = x+3$ so so $x \mapsto -3 -(x+3) = -6-x$.

And so $4 \mapsto -6 - 4 = -10$. And $-10\mapsto -6 -(-10) = 4$.

It's all good, man.

Answer 3

If $f(x)=f(c-x)$ for all $x$, then $x=\frac{c}2$ is an axis of symmetry since $$f\left(\frac{c}2+\epsilon\right)=f\left(c-\left(\frac{c}2+\epsilon\right)\right)=f\left(\frac{c}2-\epsilon\right)$$

Since $f(x)=x(6-x)=f(6-x)$, the axis of symmetry is $x=3$.

Since $f(x)=6x+x^2=-(-6x-x^2)=-x(-6-x)=f(-6-x)$, the axis of symmetry is $x=-3$.

Now, let's consider a general quadratic equation $f(x)=ax^2+bx+c$.

We have

$$f(x)=-ax\left( -\frac{b}a-x \right)+c=f\left( -\frac{b}{a}-x\right),$$

The axis of symmetry is $x=-\frac{b}{2a}$.